(a)
Interpretation:
The sample of human skin cell is whether a macroscopic, microscopic or particulate matter is to be identified.
Concept introduction:
The macroscopic system includes samples of matter those are visible with the naked eye. The microscopic system includes matter those cannot be seen by the naked eye. The microscopic matter requires an optical microscope to be seen. The particulate level includes samples of tiny particles that make up matter. The sample of matter at particulate level is too small to be seen with the most powerful optical microscope.
(b)
Interpretation:
The sample of a sugar molecule is whether a macroscopic, microscopic or particulate matter is to be identified.
Concept introduction:
The macroscopic system includes samples of matter those are visible with the naked eye. The microscopic system includes matter those cannot be seen by the naked eye. The microscopic matter requires an optical microscope to be seen. The particulate level includes samples of tiny particles that make up matter. The sample of matter at particulate level is too small to be seen with the most powerful optical microscope.
(c)
Interpretation:
The sample of a blade of grass is whether a macroscopic, microscopic or particulate matter is to be identified.
Concept introduction:
The macroscopic system includes samples of matter those are visible with the naked eye. The microscopic system includes matter those cannot be seen by the naked eye. The particulate level includes samples of tiny particles that make up matter.
(d)
Interpretation:
The sample of a helium atom is whether a macroscopic, microscopic or particulate matter is to be identified.
Concept introduction:
The macroscopic system includes samples of matter those are visible with the naked eye. The microscopic system includes matter those cannot be seen by the naked eye. The microscopic matter requires an optical microscope to be seen. The particulate level includes samples of tiny particles that make up matter. The sample of matter at particulate level is too small to be seen with the most powerful optical microscope.
(e)
Interpretation:
The sample of a single-celled plant too small to see is whether a macroscopic, microscopic or particulate matter is to be identified.
Concept introduction:
The macroscopic system includes samples of matter those are visible with the naked eye. The microscopic system includes matter those cannot be seen by the naked eye. The microscopic matter requires an optical microscope to be seen. The particulate level includes samples of tiny particles that make up matter. The sample of matter at particulate level is too small to be seen with the most powerful optical microscope.
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Chapter 2 Solutions
Introductory Chemistry: An Active Learning Approach
- Consider the photograph and illustrations of table salt. Do they include a model? Do they include a depiction of matter at the macroscopic, microscopic, and/or particulate levels? Explain your answers.arrow_forwardOn October 21, 1982, the Bureau of the Mint changed the composition of pennies (see Exercise 120). Instead of an alloy of 95% Cu and 5% Zn by mass, a core of 99.2% Zn and 0.8% Cu with a thin shell of copper was adopted. The overall composition of the new penny was 97.6% Zn and 2.4% Cu by mass. Does this account for the difference in mass among die pennies in Exercise 120? Assume the volume of the individual metals that make up each penny can be added together to give the overall volume of the penny, and assume each penny is the same size. (Density of Cu = 8.96 g/cm3; density of Zn = 7.14 g/cm3).arrow_forwardWhich of the following processes is a physical change? (a) evaporation of water; (b) combination ofhydrogen and oxygen gas to produce water; (c) dissolution of sugar in water; (d) separation of sodium chloride (table salt) into its constituentelements, sodium and chlorine; (e) combustion of sugar to produce carbon dioxide and water.arrow_forward
- Container A and Container B were filled completely with chicken powder. The total mass of 1/3 of the chicken powder in Container B and 1/11 of the chicken powder in Container A was 570 g. If 5/11 of the chicken powder in Container A was poured out, the total mass of the chicken powder in both containers became 2.28 kg. How much chicken powder was in (a) Container A in grams? (b) Container B in grams?arrow_forwardSoft wood chips weighing 17.2 kg are placed in an iron vessel and mixed with 150.1 kg water and 22.43 kg sodium hydroxide. A steel lid seals the vessel, which is then placed in an oven at 250°C for 6 hours. Much of the wood fiber decomposes under these conditions; the vessel and lid do not react.(a) Classify each of the materials mentioned as a substance or mixture. Subclassify the substances as elements or compounds.(b) Determine the mass of the contents of the iron vesselafter the reaction.arrow_forward7. State whether each of the following represents a chemical change or merely a physical change:(a) A few grams of sucrose (table sugar) are placed in a small beaker of deionized water; the sugar crystals “disappear,” and the liquid in the beaker remains clear and colorless.(b) A copper statue, over time, turns green.(c) When a teaspoon of baking soda (sodium bicarbonate) is placed into a few ounces of vinegar (acetic acid), volumes of bubbles (effervescence) are produced.arrow_forward
- Distilling a red liquid physically separates it into a red solid and a clear, colorless liquid. The red liquid is (A) an element. (B) a compound. (C) a pure substance. (D) a solution.arrow_forwardClassify each of the following as a pure substance, a solution,or a heterogeneous mixture: (a) a gold ingot, (b) a cup of coffee,(c) a wood plank.arrow_forwardA group of students took turns using a laboratory balanceto weigh the water contained in a beaker. The results theyreported were 111.42 g, 111.67 g, 111.21 g, 135.64 g,111.02 g, 111.29 g, and 111.42 g.(a) Should any of the data be excluded before the average is calculated?(b) From the remaining measurements, calculate the average value of the mass of the water in the beaker.(c) Calculate the standard deviation s and, from it, the 95% confidence limit.arrow_forward
- A metal object of an irregular shape has a mass of 31.5 grams. When the metal is added to water in a graduated cylinder, the volume of water increases from 5.28 mL to 7.61 mL. (A) How much water does the metal displace? (B) Calculate the density of the metal. (C) Is the metal made of platinum? Explain. (Hint: the density of platinum = 21.4 g/cm³)arrow_forwardDimensional Analysis is a way of doing numerical "book-keeping" when converting quantities or performing calculations. • When converting quantities from one unit to another, conversion factors are used. Solving with Dimensional Analysis and Multiple Units: If I am in Canada where the price of gas is $1.022 USD·L1, how much will it cost me to fill up my gas tank if I travelled 125 km? • Let's also assume that my car gets an average of 30.0 miles/gallon.arrow_forwardName the SI base units that are important in chemistry,and give the SI units for expressing the following:(a) length, (b) volume, (c) mass, (d) time, (e) temperature.arrow_forward
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